KE of rotating disc

[PeterDonis]

The proposed integral in the OP impies the following physical process: a disc of dust with uniform density in some inertial frame is spun up (totally non rigidly) under the constraints that density of dust per unit area in the inertial frame remains constant.

I'm not sure this is true. There is no spin-up process described in the OP or in the integral there--the angular velocity $\omega$ is treated as a constant, not a function of time (and there is no time in the integral anyway). As far as I can see, the constraint of uniform mass distribution only has to imply that the density is uniform in the inertial rest frame of the center of rotation, when the disk is rotating with constant angular velocity $\omega$.

 

[PAllen]

I'm not sure this is true. There is no spin-up process described in the OP or in the integral there--the angular velocity $\omega$ is treated as a constant, not a function of time (and there is no time in the integral anyway). As far as I can see, the constraint of uniform mass distribution only has to imply that the density is uniform in the inertial rest frame of the center of rotation, when the disk is rotating with constant angular velocity $\omega$.

I am simply describing a procedure to arrive at the final state starting from a rest state.

 

[PeterDonis]

I am simply describing a procedure to arrive at the final state starting from a rest state.

I understand that, I just don't see how it's relevant to what's in the OP. The OP is just describing the final state, not how it's achieved. And the final state can be analyzed on its own merits, without having to consider how it's achieved. As far as I can tell, what I said in post #31 is sufficient to apply the "uniform mass distribution" condition to analyze the final state. How things would have had to change while spinning up the disc doesn't seem to me to affect that.

 

[Thor Jackson]

TL;DR: Off the back of yesterday's thread about relativistic discs I tried to work out the KE of a rotating disc and got an odd result.

Consider a disc of radius $R$ with a uniform mass distribution and total mass $M$ rotating in its own plane. In its COM's inertial rest frame it has angular velocity $\omega$ about its center. In the obvious polar coordinates in that frame the kinetic energy of a small part of the disc at radius $r$ is $$dK=\frac{M}{\pi R^2}\left(\frac 1{\sqrt{1-\omega^2r^2}}-1\right)r\,dr\,d\phi$$in units where $c=1$. I just integrate to get the total KE of the disc. The $\phi$ integral is trivial and the $r$ one is not much harder:$$\begin{eqnarray*}K&=&\frac{2M}{R^2}\int_0^R\left(\frac 1{\sqrt{1-\omega^2r^2}}-1\right)r\,dr\\
&=&\frac{2M}{\omega^2R^2}\left(1-\sqrt{1-\omega^2R^2}-\frac 12\omega^2R^2\right)
\end{eqnarray*}$$This formula has the correct limit for $|\omega R|\ll 1$ - plugging in the Taylor expansion $\sqrt{1-x^2}\approx 1-\frac 12x^2-\frac 18x^4$ gives $K\approx\frac 14MR^2\omega^2=\frac 12I\omega^2$, as expected. But it is finite in the ultra-relativistic limit where $\omega R\rightarrow 1$ - in fact it tends to the absurdly low $M$. Clearly this is wrong, since you can chip off a flake of matter near the rim and it can have arbitrarily high kinetic energy.

What's going on? The integrand looks sensible and diverges at $|\omega r|=1$, but the integral behaves in an unexpected manner. Have I overlooked something? Just screwed up the maths? Can I not study a massive rotating disc in SR at these speeds because spacetime will be significantly curved by the energy?

I had a look for the result online. I found a lot of discussion of Ehrenfest and a lot of derivations of the KE of a spinning disc in Newtonian physics, but I couldn't find the result for the total kinetic energy in the relativistic case.

The maths of your integral is fine — what’s breaking is the physical model you’re feeding into it.

You’ve assumed:

• a rigid disc
• uniform mass density in the lab/COM frame
• a single angular velocity ω all the way out to the rim, even as v→c

All three of those are incompatible with special relativity at ultra‑relativistic rim speeds.

In SR there is no such thing as a perfectly rigid rotating disc. Born rigidity already tells you that you can’t spin up a finite‑radius disc from rest to arbitrarily high angular velocity without introducing stresses that themselves gravitate and, in the extreme, require a GR treatment.

More concretely:

• If the disc is “rigid” in its own local rest frame, then in the COM frame the tangential lengths are Lorentz‑contracted, so the mass density cannot stay uniform with radius.
• If you insist on uniform density in the COM frame, then the material is not rigid in its own rest frame and the stress–energy tensor is no longer that of “dust”; you need to include stresses, not just kinetic energy.

Your integral is effectively integrating the kinetic energy of a dust distribution with a fixed ω and fixed lab‑frame density all the way out to r=c/ω. That’s a perfectly well‑posed mathematical problem, so it’s not surprising the integral converges. What it isn’t is a physically realizable rotating disc in SR.

If you instead model a thin ring and let its tangential speed approach c, the kinetic energy of that ring does diverge as expected. The paradox only appears when you try to extend that picture to a rigid, uniformly filled disc, which SR simply doesn’t allow at arbitrarily high rim speeds.

 

[PAllen]

I understand that, I just don't see how it's relevant to what's in the OP. The OP is just describing the final state, not how it's achieved. And the final state can be analyzed on its own merits, without having to consider how it's achieved. As far as I can tell, what I said in post #31 is sufficient to apply the "uniform mass distribution" condition to analyze the final state. How things would have had to change while spinning up the disc doesn't seem to me to affect that.

It’s possibly important to understand what M means. The construction of the integral means it summed from constituent rest masses.

 

[PeterDonis]

In SR there is no such thing as a perfectly rigid rotating disc.

Yes, there is. It is perfectly possible to have a disk whose motion is rigid at a constant angular velocity $\omega$. SR does not rule that out. That is the physical model that is being analyzed in this thread.

What is not possible is to have a rigid disk with a changing angular velocity--so you can't spin up a disk from rest to some nonzero $\omega$ with the motion being rigid the whole time. But, as I said in post #33, I don't see the spin-up process being included in the analysis being done in this thread, nor does it need to be.

• If the disc is “rigid” in its own local rest frame, then in the COM frame the tangential lengths are Lorentz‑contracted, so the mass density cannot stay uniform with radius.
• If you insist on uniform density in the COM frame, then the material is not rigid in its own rest frame and the stress–energy tensor is no longer that of “dust”; you need to include stresses, not just kinetic energy.

The question of density distribution, or more generally the stress-energy tensor of the disk, is separate from the question of Born rigidity; Born rigidity is a purely kinematic property, and it is simple to show that the congruence of worldlines describing a disk rotating at constant $\omega$ is Born rigid. (This congruence is usually called the "Langevin congruence" in the literature.) Born rigidity is also frame independent, since it only depends on the kinematic decomposition of the congruence, which is done entirely in terms of covariant objects. So as you state them, these points cannot be correct.

Your point about the stress-energy tensor components possibly having to become arbitrarily large as the rim of the disk is approached, in the limit $\omega R \to 1$, might be worth expanding on with some actual math, if you can. (The Greg Egan treatment that was referenced earlier was at least one attempt at this, and might be worth looking at.)

 

[PeroK]

I claim the integral in the OP, evaluated between specified r values below R corresponding to speed c, will always be finite, and no limiting process for the either inner or outer r value will be unbounded. This contradicts your claim.

Sure, if you do the integral first and then take the limit as the velocity tends to $c$. But, if you consider the construction of the integral for velocity tending to $c$ (i.e. swapping the order you take the limits), then the definite integral does not converge.

 

[PeterDonis]

It’s possibly important to understand what M means.

I think the properties of the final state are important for that. But I don't think the spin-up process is.

The construction of the integral means it summed from constituent rest masses.

I agree that, in the final state, exactly what "uniform mass distribution in the COM frame" means, and whether the integral in the OP properly captures that, is a good question (which I don't think has been fully answered yet). But again, I don't think analyzing the spin-up process is necessary for that.

 

[Thor Jackson]

Yes, there is. It is perfectly possible to have a disk whose motion is rigid at a constant angular velocity $\omega$. SR does not rule that out. That is the physical model that is being analyzed in this thread.

What is not possible is to have a rigid disk with a changing angular velocity--so you can't spin up a disk from rest to some nonzero $\omega$ with the motion being rigid the whole time. But, as I said in post #33, I don't see the spin-up process being included in the analysis being done in this thread, nor does it need to be.


The question of density distribution, or more generally the stress-energy tensor of the disk, is separate from the question of Born rigidity; Born rigidity is a purely kinematic property, and it is simple to show that the congruence of worldlines describing a disk rotating at constant $\omega$ is Born rigid. (This congruence is usually called the "Langevin congruence" in the literature.) Born rigidity is also frame independent, since it only depends on the kinematic decomposition of the congruence, which is done entirely in terms of covariant objects. So as you state them, these points cannot be correct.

Your point about the stress-energy tensor components possibly having to become arbitrarily large as the rim of the disk is approached, in the limit $\omega R \to 1$, might be worth expanding on with some actual math, if you can. (The Greg Egan treatment that was referenced earlier was at least one attempt at this, and might be worth looking at.)

Thanks — that’s a fair clarification. I agree that a disk in uniform rotation can satisfy Born rigidity; the Langevin congruence is the standard example, and nothing in my earlier comment was meant to dispute that kinematically.

The point I was trying to get at is that once you go beyond the kinematics and actually look at the stress–energy required to support that Born‑rigid motion at high rim speeds, the situation becomes much less benign.

For a Born‑rigid rotating disk, the tangential velocity field is fixed by the congruence, but the material stresses needed to maintain that motion grow with radius. In fact, if you take the usual expression for the required circumferential stress in the local rest frame, it scales roughly like

Tϕϕ∼γ2(r) r2ω2,
so as r→c/ω, the Lorentz factor diverges and the stress does as well. That means the stress–energy tensor cannot remain finite all the way out to the light‑speed radius.

So even though the kinematic description of a rigidly rotating disk is well‑defined, the physical realization of such a disk with finite material stresses is not. Any real material would fail long before the rim approached c, and the stress–energy blow‑up is exactly what shows up when you try to compute the total energy.

That’s why the integral in the OP behaves oddly: the model being integrated is kinematically consistent but not dynamically realizable once the rim speed becomes ultra‑relativistic. The stress–energy required to maintain rigidity is what actually diverges, not the kinetic‑energy density of dust.

 

[PeroK]

The maths of your integral is fine

There's an ambiguity over which limit to take first. Even if the physical model is not valid, there's a question over the validity of the mathematics.

That said, a more sophisticated physical model might lead to a different integral and no ambiguity in the result.

 

[PeterDonis]

one can imagine that the "disc" is a swarm of many little rockets (the illustration below is generated by ChatGPT), each with its own engine, and without the attractive forces between them, but with a communication system which makes their motions synchronized. In that case there are no internal stresses at all.

But there is additional stress-energy in the problem, because the little rockets have to have fuel and rocket exhaust, and the direction of their thrust has to vary with time. So you're just substituting that extra stress-energy (which is going to be much more intractable to analyze) for the material stresses in an ordinary disk where the inter-atomic forces are what constrain the motion of each element.

 

[PeterDonis]

it scales roughly like

Please use the PF LaTeX feature to post equations. There is a LaTeX Guide link at the bottom left of each post window.

 

[PeroK]

The proposed integral in the OP impies the following physical process: a disc of dust with uniform density in some inertial frame is spun up (totally non rigidly) under the constraints that density of dust per unit area in the inertial frame remains constant.

Dust is not a continuous distribution of matter.

 

[renormalize]

Dust is not a continuous distribution of matter.

Neither is an ideal fluid, since it's made of discrete molecules that interact in a way that prevents compression and friction, yet we can usefully model such a fluid as a continuum. Why shouldn't it be possible to similarly model a system of dust particles as continuous?

 

[PeterDonis]

Dust is not a continuous distribution of matter.

It is if we are using a typical continuum model, where "dust" is just another name for "a perfect fluid with zero pressure and stress". That seems to be basically what the OP of the thread is modeling.

 

[PeroK]

Neither is an ideal fluid, since it's made of discrete molecules that interact in a way that prevents compression and friction, yet we can usefully model such a fluid as a continuum. Why shouldn't it be possible to similarly model a system of dust particles as continuous?

You can, but perhaps you can't meaningfully have the boundary of the dust cloud moving at $c$. In one sense you can, because the boundary has zero mass and zero energy. This thread is trying to make sense of the consequences of that.

 

[PeterDonis]

perhaps you can't meaningfully have the boundary of the dust cloud moving at $c$.

There is such a thing as null dust in the literature. But it's not intended to cover this kind of case.

I think @Demystifier made the point earlier that we can actually remove the boundary and just consider the open region of the disc inside it, where all the worldlines are timelike. He did it in the context of his finite model, but you could do the same thing in the continuum model. The gamma factor in the COM frame that's associated with the worldlines as the rim is approached still will increase without bound.

 

[anuttarasammyak]

But of course we want K to diverge even without adding material

In the c limit two OP disks and this disk both have mass 2M. KE of the former is finite 2M, that of the latter diverges. If we don’t provide material to the disk of mass M, it would become a ring KE of which diverges.

When density of the disk is homgeneous in IFR where COM of the rotating disk is at rest, KE remains finite.
When density of the disk is homgeneous in the rotating system where the disk is at rest, KE diveges though mass of the disk remains finite.

 

[PeterDonis]

For a Born‑rigid rotating disk, the tangential velocity field is fixed by the congruence, but the material stresses needed to maintain that motion grow with radius.

To put this another way: the proper acceleration required to keep a small piece of the disk on its circular path grows with radius, and increases without bound in the limit $\omega R \to 1$. That proper acceleration is radially inward, and its magnitude is

$$
\frac{\omega^2 r}{1 - \omega^2 r^2}
$$

This will also be the force per unit mass that is required, which in turn will drive the material stress.

There is nothing in principle in the math that forbids any of this, but physically, if the disk is made of any kind of ordinary matter that obeys energy conditions, the stress within it can't exceed its energy density in any frame, including the COM frame in which we've been analyzing the problem. That means that, for any finite $M$, there will be a finite limit to how large $\omega R$ can be without breaking the disk apart, and that finite limit will be a finite increment short of $1$.

 

[Ibix]

Thanks all. I haven't really had time to digest all the above, but it seems to be:

1. The problem statement is sloppy about various specifics - notably the meaning of the mass.

2. The maths has some traps for the unwary that got me.

3. The question of what happens at the rim at or near $c$ may be moot because other things like energy criteria may intervene.

Once again, you seem to need a lot more careful detail in the problem specification than you do in Newtonian physics. I guess all the possible interpretations of what I meant may go some way to explaining why there's no ready made answer online.

On the topic of the intent behind the problem: I was just thinking to put together a bit more maths on why you can't reach $c$ with a spinning disc, and I was surprised to find that (my naive interpretation of) my maths suggested energy considerations weren't a problem. I think the next step is to have a go at refining the problem specification and see if I can't come up with something that's not totally implausible but is still mathematically tractable.

 

[PeterDonis]

I think the next step is to have a go at refining the problem specification

One suggestion: consider the KE (or even the total energy--no need to subtract out rest energy, it just adds more math) as a function of $\omega$, holding $R$ constant, and see what happens in the limit as $\omega \to 1 / R$.

 

[PAllen]

One suggestion: consider the KE (or even the total energy--no need to subtract out rest energy, it just adds more math) as a function of $\omega$, holding $R$ constant, and see what happens in the limit as $\omega \to 1 / R$.

This would be the same as long as you have uniformly distributed rest mass per the COM inertial frame. To get any other result you have to change the mass assumption.

 

[PeterDonis]

I think the cause is that the mass distribution is specified in the inertial frame. Then, in the local frame of a rotating piece, as the angular speed increases to a rim limit of c, the local density decreases to zero. I think I came up with a similar result some years ago.

Here is how the mass distribution issue looks to me. I'm not sure if this is the same thing you're saying here or not.

The area of the area element $r dr d\phi$, in the COM inertial frame, is not constant; it increases with $r$. But the integral in the OP assigns a constant increment of the total rest mass $M$ to each area element. That means the density of rest mass, in the COM inertial frame, decreases with $r$.

The reason I'm not sure if this is the same thing that's said in the quote above is that $R$ is finite, which means the area of the area element $R dr d \phi$, i.e., on the rim of the disk, is also finite (and is independent of $\omega$--this is the area element in the COM frame, so that's expected). So the density of rest mass in an area element at the rim is also finite; it doesn't go to zero at the rim, even in the limiting case $\omega \to 1 / R$ (because that limit doesn't affect the density at all, since it's independent of $\omega$).

But I think the density varying with $r$ does mean that the term "uniform mass distribution" might not be appropriate for the way the OP defines the integral. If we want the distribution to be "uniform" in terms of rest mass density in the COM inertial frame, then the amount of rest mass per area element needs to be a function of $r$, which means the integrand will change from what's in the OP.

 

[PAllen]

Here is how the mass distribution issue looks to me. I'm not sure if this is the same thing you're saying here or not.

The area of the area element $r dr d\phi$, in the COM inertial frame, is not constant; it increases with $r$. But the integral in the OP assigns a constant increment of the total rest mass $M$ to each area element. That means the density of rest mass, in the COM inertial frame, decreases with $r$.

I think this can be dispensed with. Just keep $r d\phi$ constant along a radial path. Also, @Ibix integral in the OP already keeps rest mass density constant in the inertial frame it is posed in. The the amount of rest mass attributed to an area element is proportional to the size of the area element.

What I was getting at is if you take a local inertial frame comoving with the 'disc' at some point, its rulers are short relative to the COM frame in the tangential direction. Thus an area element of some size per the COM frame, will have its area increased by $\gamma$ as measured in a comoving frame. This means the locally observed rest mass density tends to zero for tangenial speed approaching c.

However, I later noted that this affect doesn't fully explain the behavior, because the number of elements around the rim increases by the same factor as the area decreases, so the total rest mass in a small annulus stay the same - it is invariant. My observation was correct, but ultimately irrelevant to explaining the behavior.

 

[PeterDonis]

Just keep $r d\phi$ constant along a radial path.

You can't. You don't get to freely choose how the area element works. That's already chosen for you by the fact that you're using polar coordinates in an inertial frame in flat spacetime.

If you change coordinates, then yes, in the new coordinates you could enforce a condition like this, but the rest of the integrand would also have to change.

@Ibix integral in the OP already keeps rest mass density constant in the inertial frame it is posed in.

I don't think that's true, for the reasons I gave in my last post. The rest mass per area element is constant, but the area per area element is not. And you can't change the area per area element $r dr d \phi$. It's fixed by your choice of coordinates. See above.

 

[PAllen]

You can't. You don't get to freely choose how the area element works. That's already chosen for you by the fact that you're using polar coordinates in an inertial frame in flat spacetime.

If I am picking, for example, a finite set of approximating elements, I can certainly have the angular span of each be inversely proportional to r.

I don't think that's true, for the reasons I gave in my last post. The rest mass per area element is constant, but the area per area element is not. And you can't change the area per area element $r dr d \phi$. It's fixed by your choice of coordinates. See above.

I disagree. Each area element is given a mass of $M r d\phi dr / (\pi R^2)$. Thus attributed mass is exactly proportional to the area of the element.

 

[PeterDonis]

Each area element is given a mass of $M r d\phi dr / (\pi R^2)$. Thus attributed mass is exactly proportional to the area of the element.

So $M / \pi R^2$ is the areal density. I see.

 

[PAllen]

One suggestion: consider the KE (or even the total energy--no need to subtract out rest energy, it just adds more math) as a function of $\omega$, holding $R$ constant, and see what happens in the limit as $\omega \to 1 / R$.

This is a nice simplification for further discussion, that changes the main questions not at all. This would be computing the total energy in the COM frame, which would also be the invariant mass of the disc (in natural units) computed in any frame. Then the formula becomes simply:
$$E = \frac{2M}{\omega^2R^2}\left(1-\sqrt{1-\omega^2R^2}\right)$$
The limit described is simply 2M.

There is nothing ill posed in the OP improper integral. I believe @PeroK s claim that the energy of an annulus could vary depending on how you take limits is simply wrong for this problem setup (specifically, the mass distribution). Specifically, the energy of an annulus would be finite in the limit, no matter how you took it, I will show this in my next post.

 

[PAllen]

To decouple an annulus from any hypothetical containing disc, we introduce $\rho$ which is simply $M/\pi r^2$ in post #1 convention. Then the energy of an arbitrary annulus is:
$$E=\frac{2\pi\rho}{\omega^2}\left(\sqrt{1-\omega^2 r_1^{~2}}-\sqrt{1-\omega^2 r_2^{~2}}\right)$$

There is no way to get any unbounded result from this with any valid limiting procedure. Note, the convention is $r_1 < r_2$.

 

[pervect]

Do you mean a hoop with zero or non-zero extension in the radial direction? In the case of a realistic non-zero extension, it's not easier at all. In the case of zero extension the KE will be infinite for $\omega R=1$, which does not help to understand why the result is finite for the disc.

I think it's a red herring which makes the analysis unnecessarily complicated. First, it depends on elastic properties of the material, so it misses the essential purely kinematic effect that does not depend on the material. Second, one can imagine that the "disc" is a swarm of many little rockets (the illustration below is generated by ChatGPT), each with its own engine, and without the attractive forces between them, but with a communication system which makes their motions synchronized. In that case there are no internal stresses at all.

The system with rockets isn't a closed system, so I don't think it would be a good model at all. I'd question whether a non-closed system even had a definite invariant mass or energy, due to the relativity of simultaneity. I wouldn't have any objection to a system where the rockets exhcanged particles between themselves to hold themselves in position, though I've never seen anyone write about such a thing. Of course, creating the attractive force you'd need to hold the rockets in a ring-like orbit would seem to require exchanging particles (real or virtual) of negative mass, as exchanging particles of positive mass would create forces of the wrong sign. The same if the rockets exchanged particles with some central point in the origin, you still need negative mass particles. Having the rocket exhaust move outwards does keep the rockets in the correct orbit, but at the expense of making the system not a closed one.

The covariant entity that describes the flows of energy and momentum through space-time is the stress-energy tensor.

Tension is one of the forces of stress that's part of the stress energy tensor. In most applications of General Relativity tension is irrelevant. The only terms that contribute to the stress-energy tensor in the systems usually studied with GR are positive pressures, usually isotropic.

note: Latex rendering problem, stuff was disappearing on me until I stopped triggering Latex.

A simple boost of the stress-energy tensor
T'^{ab} = \Lambda^a{}_{c} \Lambda^b{}_{d} T^{cd}
illustrates how pressure terms in one frame can contribute to energy density in another frame.

Example: suppose the only nonzero component of the stress-energy tensor T is $T^{33$. Then $\Lambda^0{}_3 \Lambda^0{}_3$ would make that term contribute to the $T'^{00}$ term in the "boosted" stress-energy tensor T'.

I don't think it's at all unreasonable to ask in the hoop problem what the stress-energy tensor of the hoop is. I would (and did) go so far as to say that the stress-energy tensor is the only covariant mathematical description of the hoop. I should add - that I'm aware of. Every time I say something is the only way, I find out that there is actually another way. Be that as it may, I don't know any other way that's covariant to describe the distribution of mass-energy.

While the stress energy tensor is mostly used in GR, it originates in SR. That said, it's usually only brought up in the context of GR, the only author I've read that talks about it in a SR context is Rindler, and I don't particularly care for his treatmen.